Computing the Killing form on \(\mathfrak{so}(n)\) and \(\mathfrak{su}(n)\):
The Cartan-Killing form \[K : \mathfrak{g} \times \mathfrak{g} \to \mathbb{R}\] of the Lie algebra \(\mathfrak{g}\) of a Lie group \(G\) is defined by \[K(X,Y) \overset{\operatorname{df}}{=} \operatorname{tr}(\operatorname{ad}_X \circ \operatorname{ad}_Y),\] and arises as a natural definition of an \(\left(\operatorname{Ad} : G \to \operatorname{Aut}(\mathfrak{g})\right)\)-invariant inner product on \(\mathfrak{g}\). It can be computed as follows:
First, note that since it’s an invariant inner product, the following claim shows that it’s a scalar multiple of the usual inner product i.e. trace form on \(\mathfrak{g}\) (since the Killing form is invariant under either action, the adjoint representation of \(\operatorname{SU}(n)\) is irreducible for all \(n\) and the adjoint representation of \(\operatorname{SO}(n)\) is irreducible for all \(n\) except \(n = 4\), but then \(\mathfrak{so}(4) \simeq \mathfrak{su}(2) \oplus \mathfrak{su}(2)\)):
Claim.
Let \(K\) be a compact simple Lie group such that \(K\) admits an irreducible representation on the real vector space \(V\). Then any two inner products invariant under the \(K\)-action are proportional by a constant.
Proof of claim.
If there are two inner products on \(V\), say \(\phi(x,y)\) and \(\psi(x,y)\), then there is a \(T\) such that \((V, \phi) \simeq (V,\psi)\) as inner product spaces over \(\mathbb{R}\), i.e. \(\phi\) and \(\psi\) are equal to a twist by \(T\): \[\phi(x,y) = \psi(x,Ty).\] If \(\phi\) and \(\psi\) are \(K\)-invariant, then \(T\) must be \(K\)-equivariant. Since \(V\) was an irreducible representation, it follows from Schur’s lemma that every \(K\)-equivariant operator is a multiple of the identity, hence by bilinearity there is a \(c\) such that \[\phi(x,y) = c \psi(x,y).\]
So it remains to compute this constant \(c_n\) for \(\mathfrak{so}(n)\) and \(\mathfrak{su}(n)\). Since we know \(c\) is uniform in arguments for \(K\), we can just compute it for a specific choice of \(X\) and \(Y\). \(c_1\) is \(0\) for either \(\mathfrak{su}(1)\) or \(\mathfrak{so}(1)\) because then the Lie bracket is trivial. Let’s pick \(X = Y = B_{1,2}\) the matrix with \(1\) in the \((1,2)\)th entry, \(-1\) in the \((2,1)\)th entry, and \(0\)’s elsewhere, which is always a basis element of \(\mathfrak{su}(n)\) and \(\mathfrak{so}(n)\) for \(n \geq 2\). \(\operatorname{tr}\left(B_{1,2} B_{1,2}\right)\) is easily seen to be \(-2\).
Now, since \(\operatorname{ad}(X)\) is \([X,-]\), \(K(B_{1,2}, B_{1,2})\) is seen to simplify to the expression \[\operatorname{tr}\left(Z \mapsto \left(\left(B_{1,2}\right)^2 Z + Z \left(B_{1,2}\right)^2 - 2 B_{1,2} Z B_{1,2}\right)\right)\] and factoring out \(-1\) yields \[= \operatorname{tr}\left(Z \mapsto -\left(\begin{pmatrix} 1& & \\\ &1 & \\\ & & 0 \dots \end{pmatrix}Z + Z \begin{pmatrix} 1& & \\\ &1 & \\\ & & 0 \dots \end{pmatrix} - 2\begin{pmatrix} & -1& \\\ 1& & \\\ & & \end{pmatrix} Z\begin{pmatrix} &-1 & \\\ 1& & \\\ & & \end{pmatrix}\right)\right)\] which simplifies (writing \(Z = (z_{i,j})_{i,j}\)) to \[= - \operatorname{tr} \left(Z \mapsto \begin{pmatrix} 2(z_{1,1} - z_{2,2}) & 2(z_{1,2} + z_{2,1}) & z_{1,3} & \cdots& z_{1,n} \\\ 2(z_{2,1} + z_{1,2})& 2(z_{2,2} - z_{1,1}) & z_{2,3} & \cdots &z_{2,n} \\\ z_{3,1} &z_{3,2} &0 & \cdots& 0 \\\ \vdots & \vdots & \vdots& \ddots&\vdots \\\ z_{n,1}& z_{n,2}& 0& \cdots &0 \end{pmatrix}\right).\] Now, the thing enclosed in the trace above in particular sends \(B_{1,2}\) to \(0\); in fact it sends all but \(2(n-2)\) (corresponding to the entries of the first two rows outside of the first \(2 \times 2\) block, so it leaves these unchanged) of the other basis elements \(\{B_{i,j} \operatorname{\big{|}} i < j \leq n\}\) to zero, so in the case of \(\mathfrak{so}(n)\) the thing enclosed in the trace above has matrix representation given a diagonal matrix with \(2(n-2)\) \(1\)s on the diagonal. Therefore, in the case of \(\mathfrak{so}(n)\): \[K(B_{1,2}, B_{1,2}) = -2(n-2) \implies c_{\mathfrak{so}(n)} = \dfrac{K(B_{1,2},B_{1,2})}{\operatorname{tr}\left(\left(B_{1,2}\right)^2\right)} = \dfrac{-2(n-2)}{-2} = n-2.\] Similar considerations apply to \(\mathfrak{su}(n)\), which is generated by the \(n^2 - 1\) generalized Gell-Mann matrices, which is given by the collection \(\{\lambda_{i,j}^R, \lambda_{i,j}^I, \lambda_{\ell}^D\}\), as in
(Antisymmetric in the real part)
\(\lambda^R_{i,j} \overset{\operatorname{df}}{=} B_{i,j}\) for \(1 \leq i < j \leq n\) as above,
(Symmetric in the imaginary part)
\(\lambda^{I}_{i,j} \overset{\operatorname{df}}{=} i(E_{i,j} + E_{j,i})\) for \(1 \leq i < j \leq n\) where \(E_{i,j}\) is \(1\) in the \(i,j\)th place and \(0\) elsewhere, and
(Diagonal)
\(\lambda^{D}_{\ell} \overset{\operatorname{df}}{=} \sqrt{\dfrac{2}{\ell(\ell + 1)}} \cdot i \left(\left(\sum_{j = 1}^{\ell} E_{j,j}\right) - \ell E_{(\ell + 1),(\ell + 1)}\right)\) for \(1 \leq \ell \leq n -1\).
Now, how does the thing enclosed in the trace above behave on these basis elements of \(\mathfrak{su}(n)\)? Well, it again kills all but \(2(n - 2)\) of the \(\lambda^R_{i,j}\). It also kills all but \(2(n - 2)\) of the \(\lambda^I_{i,j}\). So these contribute \(4(n - 2)\) to the trace. What about the diagonal Gell-Mann matrices? Well, for \(\ell > 1\), the thing enclosed in the trace above kills \(\lambda^{D}_{\ell}\) and for \(\ell = 1\), \(\lambda^{D}_{1}\) gets sent to \[\begin{pmatrix} 4i& 0 & 0 & \cdots \\\ 0& 4i& 0& \cdots \\\ \vdots& \vdots &\ddots & \\\ \end{pmatrix}\] so the diagonal Gell-Mann matrices \(\lambda^{D}_{\ell}\) contribute \(8\) to the trace. Hence, for the case of \(\mathfrak{su}(n)\), \[K(B_{1,2}, B_{1,2}) = -4(n - 2) - 8\]\[\implies c_{\mathfrak{su}(n)} = \dfrac{K(B_{1,2},B_{1,2})}{\operatorname{tr}\left(\left(B_{1,2}\right)^2\right)} = \dfrac{-4(n - 2) - 8}{-2} = 2n - 4 + 4 = 2n.\]
